Chapter 3
Dirichlet series and arithmetic
functions
An arithmetic function is a function f : Z>0 → C. To such a function we associate
its Dirichlet series
∞
X
Lf (s) =
f (n)n−s
n=1
where s is a complex variable. It is common practice (although this doesn’t make
sense) to write s = σ + it, where σ = Re s and t = Im s. It has shown very fruitful
in number theory, to study an arithmetic function by means of its Dirichlet series.
In this chapter, we prove some basic properties of Dirichlet series and arithmetic
functions.
3.1
Dirichlet series
We want to develop a theory for Dirichlet series similar to that for power series.
P
n
Every power series ∞
n=0 an z has a radius of convergence R such that the series
converges if |z| < R and diverges if |z| > R. As we will see, a Dirichlet series Lf (s) =
P∞
−s
has an abscissa of convergence σ0 (f ) such that the series converges
n=1 f (n)n
for all s ∈ C with Re s > σ0 (f ) and diverges for all s ∈ C with Re s < σ0 (f ). For
P
−s
instance, ζ(s) = ∞
has abscissa of convergence 1.
n=1 n
We start with an important summation result, which we shall use very frequently.
59
Theorem 3.1 (Partial summation, summation by parts). Let M, N be reals with
M < N . Let x1 , . . . , xr be real numbers with M 6 x1 < · · · < xr 6 N , let
P
a(x1 ), . . . , a(xr ) be complex numbers, and put A(t) := xk 6t a(xk ) for t ∈ [M, N ].
Further, let g : [M, N ] → C be a differentiable function. Then
r
X
Z
N
A(t)g 0 (t)dt.
a(xk )g(xk ) = A(N )g(N ) −
M
k=1
Proof. Let x0 < M and put A(x0 ) := 0. Then
r
X
a(xk )g(xk ) =
k=1
=
r
X
k=1
r
X
(A(xk ) − A(xk−1 ))g(xk )
A(xk )g(xk ) −
r−1
X
k=1
A(xk )g(xk+1 )
k=1
= A(xr )g(xr ) −
r−1
X
A(xk )(g(xk+1 ) − g(xk )).
k=1
Since A(t) = A(xk ) for xk 6 t < xk+1 we have
Z
xk+1
A(xk )(g(xk+1 ) − g(xk )) =
A(t)g 0 (t)dt.
xk
Hence
(3.1)
r
X
a(xk )g(xk ) = A(xr )g(xr ) −
r−1 Z
X
Zk=1xr
k=1
= A(xr )g(xr ) −
xk+1
A(t)g 0 (t)dt
xk
A(t)g 0 (t)dt.
x1
In case that x1 = M ,xr = N we are done. if x1 > M , then A(t) = 0 for M 6 t < x1
Rx
and thus, M1 A(t)g 0 (t)dt = 0. If xr < N , then A(t) = A(xr ) for xr 6 t 6 N , hence
Z
N
A(t)g 0 (t)dt = A(N )g(N ) − A(xr )g(xr ).
xr
Together with (3.1) this implies our Theorem.
60
Theorem 3.2. Let f : Z>0 → C be an arithmetic function with the property that
P
there exists a constant C > 0 such that | N
n=1 f (n)| 6 C for every N > 1. Then
P∞
−s
Lf (s) = n=1 f (n)n converges for every s ∈ C with Re s > 0.
More precisely, on {s ∈ C : Re s > 0} the function Lf is analytic, and for its k-th
derivative we have
∞
X
(k)
Lf (s) =
f (n)(− log n)k n−s .
n=1
Proof. Notice that on {s ∈ C : Re s > 0}, the partial sums
Lf,N (s) :=
N
X
f (n)n−s (N = 1, 2, . . .)
n=1
P
(k)
k −s
are analytic, and Lf,N (s) = N
for k > 0. We have to show
n=1 f (n)(− log n) n
that the partial sums converge on {s ∈ C : Re s > 0}, and that analyticity and the
formula for the k-th derivative are maintained if we let N → ∞.
Let s ∈ C, Re s > 0. We first rewrite Lf,N (s) using partial summation. Let
P
F (t) := 16n6t f (n). By Theorem 3.1 (with {x1 , . . . , xr } = {1, . . . , N } and g(t) =
t−s ) we have
Z N
Z N
−s
−s−1
−s
Lf,N (s) = F (N )N −
F (t)(−s)t
dt = F (N )N + s
F (t)t−s−1 dt.
1
1
By assumption, there is C > 0 such that |F (t)| 6 C for every t > 1. Further
|t−s−1 | = t−Re s−1 . Hence |F (t)t−s−1 | 6 Ct−Re s−1 . Since Re s > 0, the integral
R ∞ −Re s−1
R∞
t
dt
converges,
therefore,
F (t)t−s−1 dt converges. Further, |F (N )N −s | 6
1
1
C · N −Re s → 0 as N → ∞. It follows that Lf (s) = limN →∞ Lf,N (s) converges if
Re s > 0.
We apply Theorem 2.24 to the sequence of partial sums {Lf,N (s)}. Let K be a
compact subset of {s ∈ C : Re s > 0}. There are σ > 0, A > 0 such that Re s > σ,
|s| 6 A for s ∈ K. Thus, for s ∈ K and N > 1, we have
Z N
−s
|Lf,N (s)| 6 |F (N )N | + |s|
|F (t)t−s−1 |dt
1
6 C · N −σ + A
Z
N
C · t−σ−1 dt = C · N −σ + AC · σ −1 (1 − N −σ )
1
−1
6 C + AC · σ ,
61
which is an upper bound independent of s, N .
Now Theorem 2.24 implies that for s ∈ C with Re s > 0, the series Lf (s) =
limN →∞ Lf,N (s) is analytic and moreover,
(k)
Lf (s)
= lim
N →∞
(k)
Lf,N (s)
=
∞
X
f (n)(− log n)k n−s .
n=1
Corollary 3.3. Let f : Z>0 → C be an arithmetic function and let s0 ∈ C be such
P
−s0
that ∞
converges. Then for s ∈ C with Re s > Re s0 the function Lf
n=1 f (n)n
converges and is analytic, and
(k)
Lf (s) =
∞
X
f (n)(− log n)k n−s for k > 1.
n=1
Proof. Write s = s0 + s0 . Then Re s0 > 0 if Re s > Re s0 . There is C > 0 such that
P
P
0
−s0
−s0
)n−s .
| 6 C for all N . Apply Theorem 3.2 to ∞
| N
n=1 (f (n)n
n=1 f (n)n
Theorem 3.4. There exists a number σ0 (f ) with −∞ 6 σ0 (f ) 6 ∞ such that
Lf (s) converges for all s ∈ C with Re s > σ0 (f ) and diverges for all s ∈ C with
Re s < σ0 (f ).
Moreover, if σ0 (f ) < ∞, then for s ∈ C with Re s > σ0 (f ) the function Lf is
analytic, and
(3.2)
(k)
Lf (s) =
∞
X
f (n)(− log n)k n−s for k > 1.
n=1
Proof. If there is no s ∈ C for which Lf (s) converges we have σ0 (f ) = ∞. Assume
that Lf (s) converges for some s ∈ C and define
σ0 (f ) := inf σ : ∃s ∈ C such that Re s = σ, Lf (s) converges .
Clearly, Lf (s) diverges if Re s < σ0 (f ). To prove that Lf (s) converges for Re s >
σ0 (f ), take such s and choose s0 such that σ0 (f ) < Re s0 < Re s and Lf (s0 ) con(k)
verges. By Corollary 3.3, Lf is convergent and analytic in s, and for Lf (s) we have
expression (3.2).
62
The number σ0 (f ) is called the abscissa of convergence of Lf .
There exists also a real number σa (f ), called the abscissa of absolute convergence
of Lf such that Lf (s) converges absolutely if Re s > σa (f ), and does not converge
absolutely if Re s < σa (f ).
In fact, we have σa (f ) = σ0 (|f |), that is the abscissa of convergence of L|f | (s) =
P∞
P∞
P∞
−s
−s
−σ
conn=1 |f (n)|n . For write σ = Re s. Then
n=1 |f (n)n |=
n=1 |f (n)|n
verges if σ > σ0 (|f |) and diverges if σ < σ0 (|f |).
Theorem 3.5. For every arithmetic function f : Z>0 → C we have σ0 (f ) 6 σa (f ) 6
σ0 (f ) + 1.
Proof. It is clear that σ0 (f ) 6 σa (f ). To prove σa (f ) 6 σ0 (f ) + 1, we have to show
that Lf (s) converges absolutely if Re s > σ0 (f ) + 1.
Take such s; then Re s = σ0 (f ) + 1 + ε with ε > 0. Put σ := σ0 (f ) + ε/2. The
P
−σ
converges, hence there is a constant C such that |f (n)n−σ | 6 C
series ∞
n=1 f (n)n
for all n. Therefore,
|f (n)n−s | = |f (n)| · n−Re s = |f (n)n−σ | · n−1−ε/2 6 Cn−1−ε/2
P∞
P
−s
−1−ε/2
n
converges,
hence
for n > 1. The series ∞
n=1 |f (n)n | converges.
n=1
Exercise 3.1. Show that there exist arithmetic functions f such that σa (f ) =
σ0 (f ) + 1.
The next theorem implies that an arithmetic function is uniquely determined by
its Dirichlet series.
Theorem 3.6. Let f, g : Z>0 → C be two arithmetic functions for which there is
σ ∈ R such that Lf (s), Lg (s) converge absolutely and Lf (s) = Lg (s) for all s ∈ C
with Re s > σ. Then f = g.
Proof. Let h := f − g. Our assumptions imply that Lh (s) converges absolutely, and
Lh (s) = 0 for all s ∈ C with Re s > σ. We have to prove that h = 0.
Assume that there are positive integers n with h(n) 6= 0, and let m be the
smallest such n. Then for all s ∈ C with Re s > σ we have
h(m)m−s = −
∞
X
n=m+1
63
h(n)n−s .
Let σ1 > σ, and let s ∈ C with Re s > σ1 . Then
|h(m)| 6
∞
X
|h(n)|(m/n)
Re s
=
n=m+1
6 mσ1
∞
X
|h(n)|(m/n)σ1 (m/n)Re s−σ1
n=m+1
∞
X
!
|h(n)| · n−σ1
· (m/(m + 1))Re s−σ1 .
n=m+1
The series between the parentheses is convergent, hence a finite number. So the
right-hand side tends to 0 as Re s → ∞. This contradicts that h(m) 6= 0.
3.2
Arithmetic functions
A multiplicative function is an arithmetic function f such that f 6≡ 0 and f (mn) =
f (m)f (n) for all positive integers m, n with gcd(m, n) = 1. A strongly multiplicative
function is an arithmetic function f with the property that f 6≡ 0 and f (mn) =
f (m)f (n) for all integers m, n.
Notation. In expressions pk11 · · · pkt t it is always assumed that the pi are distinct
prime numbers, and the ki positive integers.
We start with some simple observations.
Lemma 3.7. (i) Let f be a multiplicative function. Then f (1) = 1. Further, if
n = pk11 · · · pkt t , then f (n) = f (pk11 ) · · · f (pkt t ).
(ii) Let f, g be two multiplicative functions such that f (pk ) = g(pk ) for every prime
p and k ∈ Z>1 . Then f = g.
(iii) let f, g be two strongly multiplicative functions such that f (p) = g(p) for every
prime p. Then f = g.
Proof. Obvious.
We define the convolution product f ∗ g of two arithmetic functions f, g by
X
(f ∗ g)(n) :=
f (n/d)g(d) for n ∈ Z>0 ,
d|n
where 0 d | n0 means that the sum is taken over all positive divisors of n.
64
Examples. Define the arithmetic functions e, E by
e(1) = 1, e(n) = 0 for all n ∈ Z>1 ,
E(n) = 1 for all n ∈ Z>0 .
Clearly, e is multiplicative, and E is strongly multiplicative. If f is any arithmetic
function, then e ∗ f = f , while
(E ∗ f )(n) =
X
f (d).
d|n
Lemma 3.8. (i) For any two arithmetic functions f, g we have f ∗ g = g ∗ f .
(ii) For any three arithmetic functions f, g, h we have (f ∗ g) ∗ h = f ∗ (g ∗ h).
Proof. Straightforward.
Theorem 3.9. (i) Let A be the set of arithmetic functions f with f (1) 6= 0. Then
A with ∗ is an abelian group with unit element e.
(ii) Let M be the set of multiplicative functions. Then M with ∗ is a subgroup of
A.
Proof. (i) We know already that ∗ is commutative and associative and that e is the
unit element of ∗. It remains to verify that every element of A has a (necessarily
unique) inverse with respect to ∗. Let f ∈ A. Notice that for an arithmetic function
g we have
f ∗ g = e ⇐⇒ f (1)g(1) = 1,
X
f (n/d)g(d) = 0 for n > 1
d|n
⇐⇒ g(1) := f (1)−1 ,
g(n) := −f (1)−1
X
f (n/d)g(d) for n > 1.
d|n, d<n
Clearly, the function g can be defined inductively by these last two relations. This
shows that f has an inverse with respect to ∗.
(ii) We first have to verify that the convolution product of two multiplicative
functions is again multiplicative. Here we use that if m, n are two coprime integers
and d is a positive divisor of mn, then d has a unique decomposition d = d1 d2 where
65
d1 is a positive divisor of m and d2 a positive divisor of n. Now let f, g ∈ M and
let m, n be two coprime positive integers. Then
X
X
(f ∗ g)(mn) =
f (mn/d)g(d) =
f (mn/d1 d2 )g(d1 d2 )
d1 |m, d2 |n
d|mn
=
XX
f (m/d1 )f (n/d2 )g(d1 )g(d2 )
d1 |m d2 |n

 

X
X
= 
f (m/d1 )g(d1 ) · 
f (n/d2 )g(d2 )
d1 |m
d2 |n
= (f ∗ g)(m) · (f ∗ g)(n).
This shows that f ∗ g ∈ M.
It remains to show that the inverse of a multiplicative function is again multiplicative. Let f ∈ M and let f −1 be its inverse with respect to ∗. Define h
by
h(pk ) := f −1 (pk ) for any prime power pk ,
h(n) := h(pk11 ) · · · h(pkt t ) if n = pk11 · · · pkt t .
Then h is multiplicative, and (f ∗ h)(pk ) = (f ∗ f −1 )(pk ) = e(pk ) for every prime
power pk . Both f ∗ h and e are multiplicative, so in fact f ∗ h = e, h = f −1 . Hence
f −1 is multiplicative.
Example. The Möbius function µ is the inverse under ∗ of E, where E(n) = 1 for
all n.
Lemma 3.10. We have
(
(−1)t if n = p1 · · · pt with p1 , . . . , pt distinct primes,
µ(n) =
0
if n is divisible by the square of a prime.
Proof. We first compute µ at the prime powers. First, µ(1) = 1. Further, for every
prime p and positive integer k one has
X
0 = e(pk ) =
E(pk /d)µ(d) = µ(1) + µ(p) + · · · + µ(pk ).
d|pk
66
From these relations one reads off that µ(p) = −1 and µ(p2 ) = µ(p3 ) = · · · = 0.
The expression for µ(n) for arbitrary positive integers n follows by using that µ is
multiplicative.
Theorem 3.11 (Möbius’ Inversion Formula). Let f be an arithmetic function. DeP
fine F (n) := d|n f (n) for n ∈ Z>0 . Then
f (n) =
X
µ(n/d)F (d) for n ∈ Z>0 .
d|n
Proof. We have F = E ∗ f . Hence
µ ∗ F = µ ∗ (E ∗ f ) = (µ ∗ E) ∗ f = e ∗ f = f.
Examples. 1) Define ϕ(n) := #{k ∈ Z : 1 6 k 6 n, gcd(k, n) = 1}. It is
P
well-known that d|n ϕ(d) = n for n ∈ Z>0 . This implies that
ϕ(n) =
X
µ(n/d)d,
d|n
or ϕ = µ ∗ I1 , where we define Iα (n) = nα for n ∈ Z>0 , α ∈ C. As a consequence, ϕ
is multiplicative, and for n = pk11 · · · pkt t we have
ϕ(n) =
t
Y
ϕ(pki i )
t
Y
=
(pki i − piki −1 ).
i=1
i=1
P
2) Let α ∈ C and define σα (n) = d|n dα for n ∈ Z>0 . Then σα = E ∗ Iα , which
implies that σα is multiplicative. Hence for n = pk11 · · · pkt t we have
 t
Y pα(ki +1)−1


i

if α 6= 0,

t
α

Y
p
−
1
i
ki
i=1
σα (n) =
σα (pi ) =
t
 Y

i=1

(ki + 1)
if α = 0.


i=1
67
3.3
Convolution product vs. Dirichlet series
We investigate the relation between the convolution product of two arithmetic functions and their associated Dirichlet series.
Theorem 3.12. Let f, g be two arithmetic functions. Let s ∈ C be such that Lf (s)
and Lg (s) converge absolutely.
Then also Lf ∗g (s) converges absolutely, and Lf ∗g (s) = Lf (s)Lg (s).
Proof. Since both Lf (s) and Lg (s) are absolutely convergent we can rearrange their
product as a double series and then rearrange the terms:
∞
X
−s
f (m)m
∞
X
m=1
=
−s
g(n)n
n=1
∞ X
∞
X
−s
f (m)g(n)(mn)
=
m=1 n=1
∞ X
X
k=1
f (m)g(n) k −s
mn=k
∞
X
(f ∗ g)(k)k −s = Lf ∗g (s).
=
k=1
We now show that Lf ∗g (s) converges absolutely:
∞
X
−s
|(f ∗ g)(k)k | 6
k=1
=
∞ X
X
|f (m)| · |g(n)| · |k −s |
k=1
mn=k
∞
X
∞
X
|f (m)m |
|g(n)n−s | < ∞
−s
m=1
n=1
by following the above reasoning in opposite direction and taking absolute values
everywhere. This completes our proof.
P
P
Q
Q
We define p (· · · ) = limN →∞ p6N (· · · ), and p (· · · ) = limN →∞ p6N (· · · )
where the sums and products are taken over the primes.
Theorem 3.13. Let f be a multiplicative function. let s ∈ C be such that Lf (s) =
P∞
−s
converges absolutely. Then
n=1 f (n)n
(3.3)
Lf (s) =
∞
YX
p
j=0
68
f (pj )p−js
and the right-hand side converges absolutely.
P
j −js
Further, Lf (s) 6= 0 as soon as ∞
6= 0 for every prime p.
j=0 f (p )p
Remark. The absolute convergence of the right-hand side means that for every
prime p, the series
∞
X
Ap (s) :=
|f (pj )p−js |
j=0
converges, and that the infinite product
Q
p
Ap (s) converges.
Proof. We first show the absolute convergence of the right-hand side of (3.3). First
notice that for every prime p,
Ap (s) =
∞
X
j
|f (p )p
−js
|6
∞
X
|f (n)n−s | < ∞.
n=1
j=0
P
Q
Q
Recall that p Ap (s) = p (1 + |Ap (s) − 1|) converges if p |Ap (s) − 1| converges
(see the section on infinite products in the Prerequisites). But the latter holds, since
X
|Ap (s) − 1| =
∞
XX
p
p
j
|f (p )p
−js
|6
∞
X
|f (n)n−s | < ∞.
n=2
j=1
This proves the absolute convergence of the right-hand side of (3.3).
P
j −js
. We have seen that the series Lp (s) all converge
Put Lp (s) := ∞
j=0 f (p )p
absolutely. Further,
X
|Lp (s) − 1| 6
p
∞
XX
p
|f (pj )p−js | < ∞.
j=1
Q
Hence p Lp (s) converges, which implies that the product is 0 if and only if at least
one of the factors Lp (s) is 0.
Q
It remains to prove that Lf (s) = p Lp (s). Let N > 1 and let p1 , . . . , pt be
the prime numbers 6 N . Further, let SN be the set of integers composed of prime
numbers 6 N and TN the set of remaining integers, i.e., divisible by at least one
prime > N . Since the series Lp (s) (p prime) converge absolutely, we have
Y
X
X
t s
Lp (s) =
f (pj11 ) · · · f (pjt t )(p1−j1 · · · p−j
)
=
f (n)n−s .
t
p6N
j1 ,...,jt >0
n∈SN
69
Now clearly,
∞
X
X
X
Y
f (n)n−s −
Lp (s) = f (n)n−s = f (n)n−s Lf (s) −
n=1
n∈SN
p6N
∞
X
6
n∈TN
|f (n)n−s | → 0 as N → ∞.
n=N +1
This proves (3.3).
Corollary 3.14. Let f be a strongly multiplicative function. Let s ∈ C be such that
Lf (s) converges absolutely. Then
Lf (s) =
Y
p
1
.
1 − f (p)p−s
Further, Lf (s) 6= 0.
Proof. Use that
∞
X
j
f (p )p
−js
∞
X
=
(f (p)p−s )j =
j=0
j=0
1
.
1 − f (p)p−s
Further, all factors (1 − f (p)p−s )−1 are 6= 0, hence Lf (s) 6= 0.
Examples. 1) For s ∈ C with Re s > 1 we have
ζ(s) =
∞
X
n−s =
Y
(1 − p−s )−1 (Euler)
p
n=1
P
−s
converges absolutely,
2) For s ∈ C with Re s > 1, the series Lµ (s) = ∞
n=1 µ(n)n
hence
∞
∞
X
X
−s
ζ(s)Lµ (s) =
(E ∗ µ)(n)n =
e(n)n−s = 1.
n=1
n=1
P∞
That is, ζ(s)−1 = n=1 µ(n)n−s for s ∈ C with Re s > 1. An alternative way to
prove this, is to observe that
ζ(s)−1 =
∞
∞
X
Y
YX
(1 − p−s ) =
µ(pj )p−js =
µ(n)n−s .
p
p
j=0
70
n=1
3) Recall that ϕ = µ ∗ I1 . The series LI1 (s) =
absolutely for Re s > 2. Hence
∞
X
P∞
n=1
n/ns = ζ(s − 1) converges
ϕ(n)n−s = Lϕ(s) = Lµ (s)LI1 (s) = ζ(s − 1)/ζ(s)
n=1
and Lϕ (s) converges absolutely if Re s > 2.
4) The (very important) von Mangoldt function Λ is defined by
log p if n = pk for some prime p and some k > 1,
Λ(n) =
0
otherwise.
n
Λ(n)
1
0
2
log 2
3
log 3
4
log 2
5
log 5
6
0
7
log 7
8
log 2
9
log 3
10
0
For n = pk11 · · · pkt t (unique prime factorization) we have
X
d|n
Λ(n) =
ki
t X
X
log pi =
i=1 j=1
t
X
ki log pi = log n.
i=1
Hence E ∗ Λ = log, where log denotes the arithmetic function n 7→ log n. So
Λ = µ ∗ log.
P
−s
converges absoLemma 3.15. For s ∈ C with Re s > 1, the series ∞
n=1 Λ(n)n
lutely, and
∞
X
Λ(n)n−s = −ζ 0 (s)/ζ(s).
n=1
Proof. We apply Theorem 3.12. First recall that Lµ (s) converges absolutely if Re s >
P
−s
for Re s > 1. Hence
1. Further, by Theorem 3.4, we have ζ 0 (s) = ∞
n=1 (− log n)n
∞
X
n=1
| log(n)n−s | =
∞
X
(log n)n−Re s = −ζ 0 (Re s)
n=1
converges if Re s > 1. That is, Llog (s) converges absolutely if Re s > 1. It follows
that
LΛ (s) = Lµ (s)Llog (s) = −ζ(s)−1 ζ 0 (s)
and LΛ (s) converges absolutely if Re s > 1.
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